The function $f(x)=\frac{\lambda \sin x+6 \cos x}{2 \sin x+3 \cos x}$ is increasing, if
The function $f(x)=\frac{\lambda \sin x+6 \cos x}{2 \sin x+3 \cos x}$ is increasing, if
- $\lambda>2$
- $\lambda < 4$
- $\lambda \geq 4$
- $\lambda>1$
Solution
$\begin{aligned}
& f(x)=\frac{\lambda \sin x+6 \cos x}{2 \sin x+3 \cos x} \\
& f^{\prime}(x)=\frac{\{[(2 \sin x+3 \cos x)(\lambda \cos x-6 \sin x]-[\lambda \sin x+6 \cos x)(2 \cos x-3 \sin x)]\}}{(2 \sin x+3 \cos x)^2}
\end{aligned}$
When $\mathrm{f}^{\prime}(\mathrm{x}) \geq 0$, we get
$\begin{aligned}
& {\left[\left(2 \lambda \sin \mathrm{x} \cos \mathrm{x}+3 \lambda \cos ^2 \mathrm{x}-12 \sin ^2 \mathrm{x}-18 \sin \mathrm{x} \cos \mathrm{x}\right)\right.} \\
& \left.-\left(2 \lambda \sin \mathrm{x} \cos \mathrm{x}+12 \cos ^2 \mathrm{x}-3 \lambda \sin ^2 \mathrm{x}-18 \sin \mathrm{x} \cos \mathrm{x}\right)\right] \geq 0 \\
& \therefore 3 \lambda\left(\sin ^2 \mathrm{x}+\cos ^2 \mathrm{x}\right)-12\left(\sin ^2 \mathrm{x}+\cos ^2 \mathrm{x}\right) \geq 0 \\
& \therefore 3 \lambda-12 \geq 0 \Rightarrow \lambda \geq 4
\end{aligned}$
Asked in: MHT CET 2021 (22 Sep Shift 2)
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